Orthogonal Frequency-Division Multiplexing (OFDM) is a digital multi-carrier modulation scheme, which uses a large number of closely-spaced orthogonal sub-carriers. Each sub-carrier is modulated with a conventional modulation scheme (such as quadrature amplitude modulation) at a low symbol rate, maintaining data rates similar to conventional single-carrier modulation schemes in the same bandwidth.
The primary advantage of OFDM over single-carrier schemes is its ability to cope with severe channel conditions, for example, attenuation of high frequencies at a long copper wire, narrowband interference and frequency-selective fading due to multipath, without complex equalization filters. Channel equalization is simplified because OFDM may be viewed as using many slowly-modulated narrowband signals rather than one rapidly-modulated wideband signal. Low symbol rate makes the use of a guard interval between symbols affordable, making it possible to handle time-spreading and eliminate inter-symbol interference.
In OFDM communications, scattered pilots (SP) are typically used for channel estimation and equalization. Scattered pilots are complex OFDM cells with known phase and amplitude arranged in frequency and time according to a defined pattern. The pilots are typically selected from a binary alphabet, e.g., {+1; −1} and are boosted in power compared to the data cells.
FIG. 1 shows an example of such a pattern in the form of a diagonal grid, as it is used in DVB-T (digital video broadcasting terrestrial), i.e., a digital broadcasting standard based on OFDM (cf. the ETSI Standard ETS 300 744, “Digital Broadcasting Systems for Television, Sound and Data Services; framing structure, channel coding and modulation for digital terrestrial television”). The pilots are indicated by the black circles, whereas the data cells are indicated by open circles.
The scattered pattern shown in FIG. 1 is characterized by two parameters: Ds is the distance between SPs that are adjacent along the time axis in a pilot-bearing subcarrier, and Dk is the distance between two SP-bearing subcarriers that are adjacent along the frequency axis. These two parameters are also referred to as the SP spacing in time and frequency, respectively. In this type of pattern, SPs are present in every OFDM symbol and the distance between two SPs in a symbol is DsDk. The present invention builds on such a SP pattern. In DVB-T, Ds=4 and Dk=3, as shown in FIG. 1.
Since the channel (the state of the channel) usually varies (fades) in time, due to Doppler variations, and in frequency, due to multi-path delay, the SP pattern must be dense enough to sample the channel variations along both axes as required by the sampling theorem. The Ds parameter defines the sampling along the time axis, whereas the Dk parameter defines the sampling along the frequency axis.
The channel estimation process consists of two steps. First, the channel is estimated at the SP positions by dividing the received value by the known pilot value (the reference signal). Second, the channel estimates for the other cells are computed by interpolating between the estimates at the SP positions. The interpolation is conceptually two-dimensional, but can be practically performed by first interpolating in time then in frequency, as shown in FIG. 9. Moreover, interpolation can be combined with noise reduction in order to improve the accuracy of the estimate.
In order to increase the communication reliability, multiple transmitters operating in parallel in the same frequency band can be used. This is referred to in the art as multiple-input single-output (MISO), when there is one receiver, or multiple-input multiple-output (MIMO), when there are multiple receivers. As in the single-transmitter case, channel estimation is required for coherent demodulation.
In the general MIMO case, the channel between each transmitter and each receiver must be estimated. A MIMO configuration for 4 transmitters and 2 receivers is shown in FIG. 10 as an example. We can express the received signal vector y as a function of the transmitted signal vector x and the channel matrix H as shown in the following Math. 1.
                                          [                                                                                y                    1                                                                                                ⋮                                                                                                  y                    M                                                                        ]                    ⁡                      [                                                                                h                    11                                                                    …                                                                      h                                          1                      ⁢                      N                                                                                                                    ⋮                                                  ⋱                                                  ⋮                                                                                                  h                                          M                      ⁢                                                                                          ⁢                      1                                                                                        …                                                                      h                    MN                                                                        ]                          ⁡                  [                                                                      x                  1                                                                                    ⋮                                                                                      x                  N                                                              ]                                    [                  Math          .                                          ⁢          1                ]            wherein N is the number of transmitters, e.g., N=4, and M is the number of receivers. All quantities are complex valued.
A channel estimation process is performed by each receiver independently. For channel estimation purposes the number of receivers is therefore irrelevant. The signal seen by a receiver can be written as shown in the following Math. 2.
                    y        =                              ∑                          n              =              1                        N                    ⁢                                    h              n                        ⁢                          x              n                                                          [                  Math          .                                          ⁢          2                ]            
Each receiver produces estimates of the N channel components h1, . . . , hN at each OFDM cell based on the values received at the SP locations.
Any implementation of an MISO or MIMO system based on OFDM thus has to define (i) how the scattered pilots are to be encoded so that the N channel components can be easily estimated in the receiver and (ii) how the scattered pilots are to be arranged in time and frequency.
The key idea for estimating the channel components is to employ different SPs for different transmitters. In order to be able to estimate the individual channel components, the SPs are partitioned into as many subsets as there are transmitters. All pilots belonging to a subset are multiplied by a constant coefficient that depends on the subset and the transmitter (transmit antenna). In the four-transmitter (four-transmit antenna) case, there are 16 coefficients, which can be expressed as a 4×4 matrix, as shown in FIGS. 12A and 12B. The rows correspond to the transmitters (transmit antennas) and the columns to the SP subsets.
According to FIG. 12A, a pilot that is transmitted by a transmit antenna n and belongs to subset m is multiplied by a coefficient Cmn. For example, a result of multiplying a pilot by one of the four coefficients C11 (for the transmit antenna 1), C12 (for the transmit antenna 2), C13 (for the transmit antenna 3), and C14 (for the transmit antenna 4) is used as an SP to be allocated to signals transmitted in subset 1.
Regarding the values of the coefficients, there is one necessary and sufficient condition that must be met in order to be able to separate the channel components in the receiver: the coefficient matrix must be full rank, i.e. invertible. The justification is readily apparent if we write the received values at the pilot locations as in the following Math. 3.
                              y          m                =                              (                                          ∑                                  n                  =                  1                                N                            ⁢                                                h                  n                                ⁢                                  c                  mn                                                      )                    ⁢                      p            m                                              [                  Math          .                                          ⁢          3                ]            
In the above Math. 3, pm is the original value (before multiplication) of a pilot in subset m, ym is the received value at the location of the said pilot, hn is the channel between transmitter (transmit antenna) n and receiver, and cmn, is the constant pilot coefficient for subset m and transmitter (transmit antenna) n. For simplicity the channel noise has not been considered here.
The original pilot values pm are irrelevant. Denoting the ym/pm ratio by em, the equation of the above Math. 3 can be expressed in matrix form shown in the following Math. 4.
                              [                                                                      e                  1                                                                                    ⋮                                                                                      e                  M                                                              ]                =                              [                                                                                c                    11                                                                    …                                                                      c                                          1                      ⁢                      N                                                                                                                    ⋮                                                  ⋱                                                  ⋮                                                                                                  c                                          M                      ⁢                                                                                          ⁢                      1                                                                                        …                                                                      c                    MN                                                                        ]                    ⁡                      [                                                                                h                    1                                                                                                ⋮                                                                                                  h                    N                                                                        ]                                              [                  Math          .                                          ⁢          4                ]            
The channel estimates for the channels h1 to hN can be computed by left multiplying the em estimates by the inverse of the coefficients matrix as shown in the following Math. 5.
                              [                                                                      h                  1                                                                                    ⋮                                                                                      h                  N                                                              ]                =                                            [                                                                                          c                      11                                                                            …                                                                              c                                              1                        ⁢                        N                                                                                                                                  ⋮                                                        ⋱                                                        ⋮                                                                                                              c                                              M                        ⁢                                                                                                  ⁢                        1                                                                                                  …                                                                              c                      MN                                                                                  ]                                      -              1                                ⁡                      [                                                                                e                    1                                                                                                ⋮                                                                                                  e                    M                                                                        ]                                              [                  Math          .                                          ⁢          5                ]            
As apparent from the above Math. 5, an inverse of the coefficients matrix is inevitable for channel estimation. This is the reason why the coefficients matrix must be full rank.
Although any full-rank complex matrix will do, the following two matrices are typically used in the art because of their simplicity:
Unitary Diagonal Matrix (Exists for any N)
                    [                                            1                                      0                                      0                                      0                                                          0                                      1                                      0                                      0                                                          0                                      0                                      1                                      0                                                          0                                      0                                      0                                      1                                      ]                            [                  Math          .                                          ⁢          6                ]            
Hadamard Matrix (Exists Only for N=2 or a Multiple of 4)
                    [                                                            +                1                                                                    +                1                                                                    +                1                                                                    +                1                                                                                        +                1                                                                    -                1                                                                    +                1                                                                    -                1                                                                                        +                1                                                                    +                1                                                                    -                1                                                                    -                1                                                                                        +                1                                                                    -                1                                                                    -                1                                                                    +                1                                                    ]                            [                  Math          .                                          ⁢          7                ]            
Physically, using a unitary diagonal matrix means that for antenna n only the pilots in subset n are non-zero. This means that the received values at those positions corresponding to pilots of subset n can be used for computing the estimate of channel component hn, for that position, with no further signal processing.
Using a Hadamard matrix for pilot encoding requires a multiplication with this matrix to be performed in the receiver for each cell. Such a multiplication is also referred to as a Hadamard transform. The schematic of an optimized implementation, known in the art as the fast Hadamard transform, is shown in FIG. 13.
The remaining question is how to arrange the scattered pilots in time and frequency. One option is to keep the same SP pattern like in the single-transmitter case, in which case the placement is known and only the partitioning into subsets has to be clarified. Patent Citation 1 provides one possible solution to the question of how to arrange the scattered pilots in time and frequency. According to Patent Citation 1, the scattered pilots are arranged in accordance with the same pattern as in the single-transmitter case in time and frequency, and different subsets of pilots are allocated to different pilot-bearing subcarriers. In other words, the scattered pilots are partitioned into subsets according to their subcarrier index. Thus, the N subsets of scattered pilots are evenly interleaved in frequency, i.e. along the subcarrier axis. FIGS. 2 and 3 illustrate examples of subset arrangement patterns as taught by Patent Citation 1 for two and four transmitters, respectively.
An alternative approach is known from Patent Citation 2, according to which the N subsets of scattered pilots are evenly (equally-spaced) interleaved in time, i.e. along the symbol axis. FIGS. 4 and 5 illustrate examples of subset arrangement patterns as taught by Patent Citation 2 for two and four transmitters, respectively.
Instead of keeping the pilot pattern of the single-transmitter case, pilots of the N subsets may be grouped, as indicated in FIG. 6. This approach is called “grouped interleaving” and is to be contrasted to the approach of “equally-spaced interleaving” or “even interleaving” illustrated in FIGS. 2 to 5. In the case of four transmitters (transmit antennas), there may be groups of four pilots (from subsets 1/2/3/4) or groups of two pilots (from subsets 1/2 and 3/4 for example). FIG. 6 shows the former case. Referring to FIG. 6, the groups themselves are arranged so as to be scattered in time and frequency, just like the individual scattered pilots in the single-transmitter case.
In general, interleaving N pilot subsets in one direction (time or frequency) increases the effective pilot distance in said direction by a factor of N. In order to compensate for this effect and preserve the effective distance, the physical pilot distance must be decreased by the same factor. Thus, if 4 subsets are interleaved in frequency, as in Patent Citation 1, Dk must be reduced by a factor of 4. Likewise, if the 4 subsets are multiplexed in time, as in Patent Citation 2, Ds must be reduced by the same factor.
Since the physical Dk and Ds must be integers, the effective distances Dk,eff and Ds,eff will always be multiples of 4 when interleaving the subsets in one direction only. Such a granularity of the subsets may be too coarse for some applications.